If n is a positvie integers, the value of
`E=(2n+1).^(n)C_(0)+(2n-1).^(n)C_(1)+(2n-3)^(n)C_(2)+………+1. ^(n)C_(n)2` is

Options

A. `(n+1).2^(n)`
B. `f'(2)` where `f(x)=x^(n+1)`
C. `3^(n)`
D. None of these

Answer

To solve the problem, we need to evaluate the expression:

\[ E = (2n+1) \cdot C(n, 0) + (2n-1) \cdot C(n, 1) + (2n-3) \cdot C(n, 2) + \ldots + 1 \cdot C(n, n) \]

where \( C(n, r) \) denotes the binomial coefficient "n choose r".

### Step 1: Identify the general term
The general term in the expression can be written as:
\[ E = \sum_{r=0}^{n} (2n - 2r + 1) \cdot C(n, r) \]

### Step 2: Split the summation
We can split the summation into two parts:
\[ E = \sum_{r=0}^{n} (2n + 1) \cdot C(n, r) - \sum_{r=0}^{n} 2r \cdot C(n, r) \]

### Step 3: Evaluate the first summation
The first summation can be simplified:
\[ \sum_{r=0}^{n} C(n, r) = 2^n \]
Thus,
\[ \sum_{r=0}^{n} (2n + 1) \cdot C(n, r) = (2n + 1) \cdot 2^n \]

### Step 4: Evaluate the second summation
For the second summation, we use the identity:
\[ r \cdot C(n, r) = n \cdot C(n-1, r-1) \]
Thus,
\[ \sum_{r=0}^{n} 2r \cdot C(n, r) = 2n \cdot \sum_{r=1}^{n} C(n-1, r-1) = 2n \cdot 2^{n-1} \]

### Step 5: Combine the results
Now we can combine the results from the two summations:
\[ E = (2n + 1) \cdot 2^n - 2n \cdot 2^{n-1} \]

### Step 6: Simplify the expression
We can simplify the expression:
\[ E = (2n + 1) \cdot 2^n - n \cdot 2^n = (2n + 1 - n) \cdot 2^n = (n + 1) \cdot 2^n \]

### Final Result
Thus, the value of \( E \) is:
\[ E = (n + 1) \cdot 2^n \]